TITLE: Relation between linear maps and matrices
QUESTION [8 upvotes]: I've been reading Axler's "Linear Algebra Done Right", and have learned more about linear operators/ maps, but I'd like to make sure that I understand how to properly relate this information to matrices. 
First, any $m \times n$ matrix with entries in a field $F$ uniquely determines a linear transformation $T: F^n\to F^m$ by $(x_1,...,x_n)\mapsto (\sum_{j=1}^{n}a_{1j}x_j,...,\sum_{j=1}^{n}a_{mj}x_j)$, and if $T:V\to W$ is a linear map between finite dimensional vector space $V$ and $W$ over field $F$ and we fix a basis $B_1$ in $V$ and $B_2$ in $W$, with $dim(V)=n$ and $dim(W)=m$, then $M(T)$ (function that maps $T$ to its matrix with respect to $B_1$ and $B_2$ by the procedure outlined above) is an isomorphism between $L(V,W)$ (vector space of linear maps between $V$ and $W$) and $M$ at $(m,n,F)$ (vector space of $m \times n$ matrices with entries in $F$). 
QUESTIONS
1.) If I'm given a matrix with entries in $F$, how exactly would I go about determining information about it from linear maps? For example, suppose that i'm given an $n \times n$ matrix $A$ where $n$ is odd. I know that if $T$ is an operator on an odd-dimensional vector space $V$, then $T$ has an eigenvalue. Since $T$ can be any operator on any odd-dimensional vector space, can I just pick $V=F^n$? Then I could say that $A$ represents a unique linear transformation $T:F^n\to F^n$ by the assignment above, and since $T$ has an eigenvalue, $A$ must also have (the same) eigenvalue (since $T$ and $A$ are the same transformation on $F^n$)? Furthermore, if I assume the standard basis for $F^n$, then there is no other such matrix ($m \times n$ and entries in $F$) that represents the particular operator that $A$ does, and every possible operator on $F^n$ is represented by a matrix in $M$ at $(n,n,F)$. Is this all correct? Would similar arguments apply to invertibility, similarity, etc. ?
2.) Since an $m \times n$ matrix $A$ with entries in $F$ represents a unique linear map $T: F^n\to F^m$, and since finite dimensional vector spaces with the same dimension are isomorphic, can't I also interpret $A$ as representing a unique linear map $T: V\to W$, where $V$ and $W$ are any vector spaces over $F$ s.t. $dim(V)=n$ and $dim(W)=m$?

REPLY [6 votes]: The topic can indeed create some confusion. First of all, a linear map $T\colon V\to W$ is just a function and not a matrix.
However, any $m\times n$ matrix $A$ can be used to define a linear map
$$
T_A\colon F^n\to F^m
$$
by $T_A(v)=Av$ (writing vectors in $F^n$ as columns). This is where the general concept of linear maps was born from, actually.
When $V$ and $W$ is finite dimensional, say $\dim V=n$, once we have a basis $\mathcal{B}=\{v_1;v_2;\dots;v_n\}$, we can define a bijective linear map
$$
C_{\mathcal{B}}\colon V\to F^n
$$
by
$$
C_{\mathcal{B}}(v)=
\begin{bmatrix}\alpha_1\\\alpha_2\\\vdots\\\alpha_n\end{bmatrix}
\quad\text{if and only if}\quad
v=\alpha_1v_1+\alpha_2v_2+\dots+\alpha_nv_n
$$
Given a linear map $T\colon V\to W$ and a basis $\mathcal{D}=\{w_1;w_2;\dots;w_m\}$ of $W$, we can define the matrix
$$
[T]_{\mathcal{B},\mathcal{D}}=
\begin{bmatrix}
C_{\mathcal{D}}(T(v_1)) & C_{\mathcal{D}}(T(v_2)) & \dots &
  C_{\mathcal{D}}(T(v_n))
\end{bmatrix}
$$
(where we specify the matrix by its columns) with the property that, for every $v\in V$,
$$
C_{\mathcal{D}}(T(v))=[T]_{\mathcal{B},\mathcal{D}}C_{\mathcal{B}}(v)
$$
This matrix is easily seen to be unique and, as such, it contains all the information about $T$. Computations about $T$ can be translated into computations on the associated matrix $[T]_{\mathcal{B},\mathcal{D}}$. For instance, the rank of $T$ is the same as the rank of the associated matrix and this one can be computed by row reduction or other methods.
In the particular case when $V=F^n$, $W=F^m$, $B$ is the canonical basis of $V$ and $D$ is the canonical basis of $W$, we have that
$$
C_{\mathcal{B}}(v)=v
$$
and similarly on $W$, so the relation becomes
$$
T(v)=[T]_{\mathcal{B},\mathcal{D}}v
$$
and so we prove that 

any linear map $F^n\to F^m$ is of the form $T_A$, for a unique $m\times n$ matrix $A$.

Of course, if we compute the associated matrix to $T_A$ when we choose on $V$ or $W$ bases different from the canonical basis, we'll get a matrix different (in general) from $A$. Here's where confusion may arise, but keeping well distinct the concept of linear map and associated matrix (when a choice of bases is made) will help in well understanding the matter.

An important particular case is when linear maps from $V$ to $V$ are dealt with. In this case it's quite natural to consider the same basis on the domain and the codomain (but it's not mandatory, of course). So if $T\colon V\to V$ is a linear map and $\mathcal{B}$ is a basis of $V$, we have the associated matrix
$[T]_{\mathcal{B},\mathcal{B}}=[T]_{\mathcal{B}}$ (it makes sense mentioning the basis only once) with the property that, for all $v\in V$,
$$
C_{\mathcal{B}}(T(v))=[T]_{\mathcal{B}}C_{\mathcal{B}}(v)
$$
If $v$ is an eigenvector of $T$ relative to the eigenvalue $\lambda$, we have $T(v)=\lambda v$ and so
$$
[T]_{\mathcal{B}}C_{\mathcal{B}}(v)=
C_{\mathcal{B}}(T(v))=C_{\mathcal{B}}(\lambda v)=\lambda C_{\mathcal{B}}(v)
$$
and so $w=C_{\mathcal{B}}(v)$ is an eigenvector of the matrix $[T]_{\mathcal{B}}$ relative to the eigenvalue $\lambda$. Conversely, if $[T]_{\mathcal{B}}w=\lambda w$ (and $w\ne0$), we can define $v$ as the unique vector such that $w=C_{\mathcal{B}}(v)$ and then
$$
C_{\mathcal{B}}(T(v))=C_{\mathcal{B}}(\lambda v)
$$
(just follow the path backwards), and we deduce $T(v)=\lambda v$. So the eigenvalues of $[T]_{\mathcal{B}}$ are exactly the same as the eigenvalues of $T$.
Note that the basis is irrelevant for $\lambda$ being an eigenvalue: all matrices associated to $T$ (when the same basis is chosen on the domain and the codomain) share the eigenvalues.
The eigenvectors are not shared, but the isomorphism $C_{\mathcal{B}}$ allows for getting the eigenvectors of $T$ when the eigenvectors of $[T]_{\mathcal{B}}$ are known.